The Parallelogram Law

An identity hides between the two diagonals and the two sides of a parallelogram. This scene slides the exponent $p$ and checks when that identity survives. Every norm that comes from an inner product satisfies the identity. Computing with $e_1$ and $e_2$ makes the left side $2 \cdot 2^{2/p}$ and the right side $4$. The two agree only at $p = 2$. So an $\ell^p$ norm with $p$ different from 2 cannot come from any inner product. For example, a 📐 parallelogram drawn with a ruler stays exactly the same, but the measuring ruler is swapped. If swapping the ruler alone breaks the diagonal identity, that ruler never came from an inner-product blueprint.

Examples & Exercises

Prove completely that in a real inner product space, every $x, y$ satisfies the parallelogram law $\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2$.

$$\|x+y\|^2 + \|x-y\|^2 = 2\|x\|^2 + 2\|y\|^2$$

Show proof

1. Expand the norm of the sum through the inner product. In a real inner product space $\|x+y\|^2 = \langle x+y, x+y \rangle$, and bilinearity with symmetry gives $\langle x+y, x+y \rangle = \langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2$. That is, $\|x+y\|^2 = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2$.
2. Expand the norm of the difference the same way. $\|x-y\|^2 = \langle x-y, x-y \rangle = \langle x, x \rangle - \langle x, y \rangle - \langle y, x \rangle + \langle y, y \rangle = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2$. That is, $\|x-y\|^2 = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2$.
3. Add the two identities. The cross terms cancel: $\|x+y\|^2 + \|x-y\|^2 = \left(\|x\|^2 + 2\langle x, y \rangle + \|y\|^2\right) + \left(\|x\|^2 - 2\langle x, y \rangle + \|y\|^2\right) = 2\|x\|^2 + 2\|y\|^2$. This is the parallelogram law, and the proof is complete.
4. Remark: this is the real case. The law itself also holds in complex inner product spaces by the same expansion, but the polarization identity in problem 3 below changes to a form using $\mathrm{Re}$ there — not needed here.

For $e_1 = (1,0)$ and $e_2 = (0,1)$ in $\ell^p(\mathbb{R}^2)$, compute both sides of the parallelogram law and prove completely that equality holds only when $p = 2$. Conclude that for $p \ne 2$ the $\ell^p$ norm is not induced by any inner product.

$$\|e_1+e_2\|_p^2 + \|e_1-e_2\|_p^2 = 2^{1 + 2/p}, \qquad 2\|e_1\|_p^2 + 2\|e_2\|_p^2 = 4 = 2^2$$

Show proof

1. Compute each norm first. Since $e_1 + e_2 = (1,1)$, $\|e_1+e_2\|_p = \left(|1|^p + |1|^p\right)^{1/p} = 2^{1/p}$, and squaring gives $\|e_1+e_2\|_p^2 = \left(2^{1/p}\right)^2 = 2^{2/p}$. Likewise $e_1 - e_2 = (1,-1)$, so $\|e_1-e_2\|_p = \left(|1|^p + |-1|^p\right)^{1/p} = 2^{1/p}$ and $\|e_1-e_2\|_p^2 = 2^{2/p}$.
2. Collect both sides. The left side is $\|e_1+e_2\|_p^2 + \|e_1-e_2\|_p^2 = 2^{2/p} + 2^{2/p} = 2 \cdot 2^{2/p} = 2^{1 + 2/p}$. Meanwhile $\|e_1\|_p = \|e_2\|_p = 1$, so the right side is $2\|e_1\|_p^2 + 2\|e_2\|_p^2 = 2 \cdot 1 + 2 \cdot 1 = 4 = 2^2$. Hence equality is equivalent to $2^{1 + 2/p} = 2^2$.
3. $(\Rightarrow)$ Suppose equality $2^{1 + 2/p} = 2^2$ holds. The map $t \mapsto 2^t$ is injective, so the exponents must agree: $1 + 2/p = 2$. Hence $2/p = 1$, i.e. $p = 2$.
4. $(\Leftarrow)$ Conversely, suppose $p = 2$. Then $1 + 2/p = 1 + 1 = 2$, so the left side is $2^{1 + 2/p} = 2^2 = 4$, directly equal to the right side $4$. Hence equality holds.
5. Conclusion: for $p \ne 2$ the $\ell^p$ norm violates the parallelogram law. Since problem 1 shows that every inner-product norm satisfies the law, by the Jordan-von Neumann theorem the $\ell^p$ norm for $p \ne 2$ is not induced by any inner product.

Verify completely that in a real inner product space the polarization identity $\langle x, y \rangle = \frac{\|x+y\|^2 - \|x-y\|^2}{4}$ holds.

$$\langle x, y \rangle = \frac{\|x+y\|^2 - \|x-y\|^2}{4}$$

Show proof

1. Reuse the expansions from problem 1: $\|x+y\|^2 = \|x\|^2 + 2\langle x, y \rangle + \|y\|^2$ and $\|x-y\|^2 = \|x\|^2 - 2\langle x, y \rangle + \|y\|^2$.
2. Subtract the two identities. The $\|x\|^2$ and $\|y\|^2$ terms disappear and only the cross terms survive: $\|x+y\|^2 - \|x-y\|^2 = \left(\|x\|^2 + 2\langle x, y \rangle + \|y\|^2\right) - \left(\|x\|^2 - 2\langle x, y \rangle + \|y\|^2\right) = 4\langle x, y \rangle$.
3. Divide both sides by $4$: $\frac{\|x+y\|^2 - \|x-y\|^2}{4} = \langle x, y \rangle$ — the polarization identity is verified. The inner product is completely determined by the norm alone.
4. By the Jordan-von Neumann theorem, whenever a norm satisfies the parallelogram law, this very formula defines an inner product for it (statement only).

3. Sequences